POJ 3977

最新推荐文章于 2020-04-03 12:29:45 发布

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最新推荐文章于 2020-04-03 12:29:45 发布

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原文链接：http://www.cnblogs.com/hyxsolitude/p/3642053.html

本文介绍了一种解决子集和问题的方法，通过将集合分为两个相等的部分，并为每个部分生成所有可能的子集和，之后使用二分查找来确定最小绝对值的子集和及其元素数量。

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Subset

Time Limit: 30000MS		Memory Limit: 65536K
Total Submissions: 1373		Accepted: 228

Description

Given a list of N integers with absolute values no larger than 10 ¹⁵, find a non empty subset of these numbers which minimizes the absolute value of the sum of its elements. In case there are multiple subsets, choose the one with fewer elements.

Input

The input contains multiple data sets, the first line of each data set contains N <= 35, the number of elements, the next line contains N numbers no larger than 10 ¹⁵ in absolute value and separated by a single space. The input is terminated with N = 0

Output

For each data set in the input print two integers, the minimum absolute sum and the number of elements in the optimal subset.

Sample Input

1
10
3
20 100 -100
0

Sample Output

10 1
0 2

Source

Seventh ACM Egyptian National Programming Contest

把集合分成两个 N / 2的集合，然后生成一种一个集合2 ^ (n - 1)种状态的和，对于另一个集合的所有状态的和在前一个集合中二分找到一个最接近的

并找到集合元素最小的即是所求答案

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <utility>
 6 
 7 using namespace std;
 8 
 9 
10 typedef long long ll;
11 typedef pair<ll,int> pii;
12 
13 
14 const ll INF = (1e17) + 5;
15 const  int MAX = 40;
16 
17 int N;
18 ll w[MAX];
19 pii ps[(1 << 20) + 5];
20 int cal[1 << 20],dx[4] = {-1,-2,0,1};
21 
22 ll Abs(ll x) {
23         return x > 0 ? x : -x;
24 }
25 
26 void init() {
27         for(int s = 0; s < (1 << 20); ++s) {
28                 int sum = 0;
29                 for(int i = 0; i < 20; ++i) {
30                         if(s >> i & 1) ++sum;
31                 }
32                 cal[s] = sum;
33         }
34 }
35 void solve() {
36         int n = N / 2;
37         ll sw = 0;
38         for(int s = 0; s < (1 << n); ++s) {
39                 sw = 0;
40                 for(int j = 0; j < n; ++j) {
41                         if(s >> j & 1) sw += w[j];
42                 }
43                 ps[s] = make_pair(sw,cal[s]);
44         }
45         sort(ps,ps + (1 << n));
46 
47         int n1 = N - n;
48         ll ansv = INF;
49         int anss = N;
50         for(int s = 0; s < (1 << n1); ++s) {
51                 sw = 0;
52                 for(int j = n; j < N; ++j) {
53                         if(s >> (j - n) & 1) sw += w[j];
54                 }
55                 int pos = lower_bound(ps,ps + (1 << n),make_pair(-sw,-1)) - ps;
56                 ll v = INF,t = INF;
57                 for(int i = 0; i < 4; ++i) {
58                         int id = pos + dx[i];
59                         if(id >= 0 && id < (1 << n)
60                            && (ps[id].second || s)) {
61                                 if(Abs(ps[id].first + sw) < t) {
62                                         t = Abs(ps[id].first + sw);
63                                         v = ps[id].first;
64                                 }
65                         }
66                 }
67                 pos = lower_bound(ps,ps + (1 << n),make_pair(v,-1)) - ps;
68                 if(s == 0 && ps[pos].second == 0) ++pos;
69                 if(ansv > Abs(v + sw) || ansv == Abs(v + sw) && anss > cal[s] + ps[pos].second) {
70                         ansv = Abs(v + sw);
71                         anss = cal[s] + ps[pos].second;
72                 }
73         }
74 
75         printf("%I64d %d\n",ansv,anss);
76 }
77 
78 int main()
79 {
80     freopen("sw.in","r",stdin);
81     init();
82     while(~scanf("%d",&N) && N) {
83             for(int i = 0; i < N; ++i) scanf("%I64d",&w[i]);
84             solve();
85     }
86 
87     return 0;
88 }